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The area of circle O is added to its diameter. If the circumference of
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11 Jun 2021, 06:30
11 Jun 2021, 06:30
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The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?
A) –2/pi
B) 2
C) 3
D) 4
E) 5
A) –2/pi
B) 2
C) 3
D) 4
E) 5
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Posts: 6218
Given Kudos: 136
Re: The area of circle O is added to its diameter. If the circumference of
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11 Jun 2021, 06:31
11 Jun 2021, 06:31
Carcass wrote:
The area of circle O is added to its diameter. If the circumference of circle O is then subtracted from this total, the result is 4. What is the radius of circle O?
A) –2/pi
B) 2
C) 3
D) 4
E) 5
A) –2/pi
B) 2
C) 3
D) 4
E) 5
Let r = radius of circle
Area of circle = πr²
Diameter = 2r
Circumference of circle = 2rπ
The area of circle O is added to its diameter...
We get: πr² + 2r
...If the circumference of circle O is then subtracted from this total, the result is 4
We get: πr² + 2r - 2rπ = 4
From here, we CAN just test the answer choices.
First, we can SKIP answer choice A, since the radius cannot have a negative value.
What about B (2)?
Replace r with (2) to get: π(2)² + 2(2) - 2(2)π = 4
Simplify: 4π + 4 - 4π = 4
Simplify again: 4 = 4
PERFECT!
Answer: B
ALTERNATE SOLUTION
Once we get the equation πr² + 2r - 2rπ = 4, we can also solve it algebraically.
To do so, we're going to factor the expression in parts
Here's what I mean...
We have: πr² + 2r - 2rπ = 4
Subtract 4 from both sides to get: πr² + 2r - 2rπ - 4 = 0
Rearrange terms to get: πr² - 2rπ + 2r - 4 = 0
Factor as follows: πr(r - 2) + 2(r - 2) = 0
Notice that we have (x-2) in common in both parts.
So, we can combine the parts to get: (πr + 2)(r - 2) = 0
This means that EITHER πr + 2 = 0 OR r - 2 = 0
case a: πr + 2 = 0
This means: πr = -2
Since π and r are both POSITIVE, this equation has NO SOLUTION
case b: r - 2 = 0
This tells us that r = 2
Answer: B
Re: The area of circle O is added to its diameter. If the circumference of
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16 Mar 2023, 13:05
16 Mar 2023, 13:05
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
